Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 29

$\sec0.08 \approx 1$

Let $y = sec~x$ We can find $dy$: $\frac{dy}{dx} = sec~x~tan~x$ $dy = sec~x~tan~x~dx$ When $x=0$ and $dx = 0.08$: $y = sec~0 = 1$ $dy = sec~0~tan~0~(0.08) = 0$ We can approximate $sec~(0+0.08)$: $sec~0.08 \approx sec~0+dy$ $sec~0.08 \approx 1+0$ $sec~0.08 \approx 1$